The problem provided asks to calculate the volume of carbon dioxide (CO₂) produced when 5.0 grams of propane (C₃H₈) are burned in excess oxygen at 25°C and 1 atm pressure. The balanced chemical equation for the combustion of propane is:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Steps to solve:

1. Find the molar mass of propane (C₃H₈):

   $M_{\text{C₃H₈}} = (3 \times 12.01) + (8 \times 1.008) = 44.094 \, \text{g/mol}$

2. Convert grams of propane to moles:

   $n_{\text{C₃H₈}} = \frac{5.0 \, \text{g}}{44.094 \, \text{g/mol}} \approx 0.1134 \, \text{mol}$

3. Use the stoichiometry of the balanced equation:

   From the balanced equation, 1 mole of propane produces 3 moles of CO₂. So, moles of CO₂ produced:

   $n_{\text{CO₂}} = 0.1134 \, \text{mol} \times 3 = 0.3402 \, \text{mol}$

4. Use the Ideal Gas Law to find the volume of CO₂:

   The Ideal Gas Law is:

   $PV = nRT$

   where:

   • $P = 1 \, \text{atm}$
   • $V = ?$ (Volume we need to find)
   • $n = 0.3402 \, \text{mol}$
   • $R = 0.0821 \, \text{L·atm/(mol·K)}$
   • $T = 25^\circ C = 298 \, \text{K}$

   Solving for $V$:

   $V = \frac{nRT}{P} = \frac{(0.3402 \, \text{mol})(0.0821 \, \text{L·atm/(mol·K)})(298 \, \text{K})}{1 \, \text{atm}}$

   $V \approx 8.32 \, \text{L}$

Thus, the volume of carbon dioxide produced is approximately 8.32 liters.

Would you like more details on any of these steps or have further questions?

Here are five related questions to explore:

1. How would the result change if the temperature were 50°C instead of 25°C?
2. What is the molar volume of an ideal gas at STP, and how does it relate to this problem?
3. Why is excess oxygen important in combustion reactions?
4. How does the stoichiometry of the reaction help determine the amount of products?
5. What would happen if there were a limiting amount of oxygen?

Tip: The Ideal Gas Law can be applied to many scenarios in chemistry, so it's useful to practice converting temperature to Kelvin and using this law effectively!